Chapter 5.4, Problem 26E

Solution

To find: The solution of the given equation in the interval $\left[0,2\pi \right)$ and use graph to verify the solutions.

The solutions are $x=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2}$ .

Given data:

The equation is $\sin x+\sin 3x=0$ .

Formula used:

Use the double angle identity for cosine.

  $\begin{array}{c}\cos 2u=1-2{\sin }^{2}u\\ \cos 2u=2{\cos }^{2}u-1\end{array}$

Use the double angle identity for of sine.

  $\sin 2u=2\sin u\cos u$

Calculation:

Simplify the given function by separating $3x$ and using the sine sum formula.

  $\begin{array}{c}\sin x+\sin \left(2x+x\right)=0\\ \sin x+\sin 2x\cos x+\cos 2x\sin x=0\end{array}$

Simplify the above equation by using double angle identity for sine and cosine.

  $\begin{array}{c}\sin x+2\sin x\cos x\cos x+\left(2{\cos }^{2}x-1\right)\sin x=0\\ \sin x+2\sin x{\cos }^{2}x+2{\cos }^{2}x\sin x-\sin x=0\\ 4{\cos }^{2}x\sin x=0\end{array}$

Use the zero factor property. Equate the first factor to zero.

  $\begin{array}{c}\sin x=0\\ x={\sin }^{-1}\left(0\right)\\ =0,\pi \end{array}$

Equate the second factor to zero.

  $\begin{array}{c}{\cos }^{2}x=0\\ \cos x=0\\ x={\cos }^{-1}\left(0\right)\\ x=\frac{\pi }{2},\frac{3\pi }{2}\end{array}$

Draw the graph of the given equation.

  

From the above graph, it is clear that the solutions are $x=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2}$ .

Therefore, the solutions are $x=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2}$ .